Class 12th

This happens because although they have the same temeprature constant, some of their other properties can also vary such as activation energies, surface area of the reaction, concentration of the reactant, etc. As a result, the rate of a reaction comes out to be different for the reactions because of their dependence on these factors.

L: x/1=y/0=z/-1. Let a point on L be (r,0, -r).
Direction ratio of PN is (r-1, -2, -r+1).
PN is perpendicular to L, so (r-1) (1)+ (-2) (0)+ (-r+1) (-1)=0 ⇒ 2r-2=0 ⇒ r=1. N (1,0, -1).
Let Q be (s,0, -s). Direction ratio of PQ is (s-1, -2, -s+1).
PQ is parallel to plane x+y+2z=0, so normal to plane is perp to PQ.
(s-1) (1)+ (-2) (1)+ (-s+1) (2)=0 ⇒ s-1-2-2s+2=0 ⇒ -s-1=0 ⇒ s=-1. Q (-1,0,1).
PN = (0, -2,0). PQ= (-2, -2,2).
cosα = |PN.PQ|/|PN|PQ| = 4/ (2√12) = 1/√3.

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

e? - e? - 2e³? - 12e²? + e? + 1 = 0
e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
Let y=e? y? -y? -2y³-12y²+y+1=0.
The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
Either e³? = 4e? + 1 (One Solution)
OR e³? = 1 - 3e? (One Solution)
∴ the equation has total 2 solutions.

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

This is the type of reaction which bheavies like a first order reaction inspite of being a higher order reaction (second or third). This happens due to a particular reactant being present in an excessive quantity thorughout the chemical reaction.

These are the important highlights of integrated rate equations:

• predict the rate of a reaction
• calculate order of reaction (zero, first, second)
• understand mechanism of a reaction
• compute half life
• calculate the value of k

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4